Connected linear Lie groups $G$ generated by $\exp (\mathfrak{g})$.

Question

Let $G$ be a connected linear Lie group with Lie algebra $\mathfrak{g}$. I understand that any open neighborhood of the identity of $G$ generates it, but, why does $\exp(\mathfrak{g})$ also generate it? Is $\exp(\mathfrak{g})$ open?

Answer 1

$\exp(\cdot): \mathfrak{g}\to G$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $\exp(\cdot)$ maps an open neighborhood $0\in V \subseteq \mathfrak{g}$ to an open neighborhood around the identity of $G$. Now, $H=\langle \exp(V)\rangle$ is a connected component of $G$ containing the identity. Since $G$ is connected, this subgroup must be $G$.